福建省福州市山海联盟校教学协作体2023-2024学年高一上学期期中考试数学试卷

这是一份福建省福州市山海联盟校教学协作体2023-2024学年高一上学期期中考试数学试卷，文件包含23-24上山海联盟校教学协作体期中考数学试卷pdf、23-24上山海联盟校教学协作体期中考数学答案docx等2份试卷配套教学资源，其中试卷共9页， 欢迎下载使用。
1. B2. C3. B4. D
5. D6. C7. C8. A
二、多项选择题：本大题共4个小题，每小题5分，共20分.在每小题给出的四个选项中，有多项符合题目要求.全部选对的得5分，选对但不全的得2分，有选错的得0分．
9.ABD 10. CD11. BD12. AB
三、填空题：每小题5分，其中14题，满分20分．
13.x≥0， 14.
15. QUOTE -1 16. QUOTE
四、解答题：本题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤.
17.【解析】（1）当时，，·································1分
∵，
∴；·····································4分
（2）由已知，
当时，，解得；·················6分
当时，，解得；···············9分
综上所述，，即.··········································10分
QUOTE
QUOTE
18.【解析】证明：因为··················2分
，··········································4分
所以，······································· 5分
当且仅当，时，不等式中等号成立.·························6分
（2），·················9分
当且仅当，即或时，不等式中等号成立.·················11分
所以的最小值为4.···········································12分
19.解析】（1），····························4分
所以；·············································· 6分
（2）设一次函数的解析式为，···························8分
则，解得，·······························11分
所以.···························12分
【解析】（1）由题知，····················· 2分
解得：.·············································4分
（2），， 对称轴.····················· 5分
当即时，函数在上单调增，···················· 6分
则. ············································ 7分
当即时，函数在上单调减，在上单调增，····· 8分
则.···········································9分
当即时，函数在上单调减，···························· 10分
则. ············································ 11分
综上，········································12分
【解析】(1)解：每吨平均成本为（万元）····················································1分
则，·························································4分
当且仅当，即时取等号.······························································5分
∴年产量为200吨时，每吨平均成本最低为32万元.··············································6分
解：设年获得总利润为万元，则·········································7分
·····································································8分
∵在上是增函数，······································································· 9分
∴时，有最大值为（万元）.·········11分
∴年产量为210吨时，可获得最大利润1660万元.···············································12分
22.【解析】（1）令，则, 可得；·····················2分
（2）在上单调递减，证明如下：·····························································3分
由已知，对于有成立，，
令，则，
所以，对有，故是奇函数，·············································4分
任取且，则，由已知有，······························5分
又，得··························6分
所以在上是减函数；
（3）因为，
所以，
即，··························································7分
因为在上是减函数，
所以， 即，又，
所以，················································································8分
当时，即时，原不等式的解集为；·························9分
当时，即时，原不等式的解集为；················································10分
当时，即时，原不等式的解集为.······························11分
综上所述：当时，原不等式的解集为；
当时，原不等式的解集为；
当时，原不等式的解集为.···················································12分